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3 years ago

Find value of x and y

Mathematics

1 answer:

marshall27 [118]3 years ago

8 0

Answer:

x = 17 $\sqrt{3}$ , y = 34

Step-by-step explanation:

Since the triangle is right use trig ratios to find x and y

note exact values sin30° = $\frac{1}{2}$ , tan30° = $\frac{1}{\sqrt{3} }$

sin30° = $\frac{opposite}{hypotenuse}$ = $\frac{17}{y}$

multiply both sides by y

y sin30° = 17 ( divide both sides by sin30° )

y = $\frac{17}{\frac{1}{2} }$ = 34

tan30° = $\frac{opposite}{adjacent}$ = $\frac{17}{x}$

multiply both sides by x

x tan30° = 17 ( divide both sides by tan30° )

x = $\frac{17}{\frac{1}{\sqrt{3} } }$ = 17 $\sqrt{3}$

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A survey reported in Time magazine included the question ‘‘Do you favor a federal law requiring a day waiting period to purchase

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Answer:

The 90% confidence interval for the difference between proportions is (-0.260, -0.165).

Step-by-step explanation:

The question is incomplete. The complete question is:

"A survey reported in Time magazine included the question ‘‘Do you favor a federal law requiring a 15 day waiting period to purchase a gun?” Results from a random sample of US citizens showed that 318 of the 520 men who were surveyed supported this proposed law while 379 of the 460 women sampled said ‘‘yes”. Use this information to find a 90% confidence interval for the difference in the two proportions, pm - pw. Subscript pm is the proportion of men who support the proposed law and pw is the proportion of women who support the proposed law. (Round answers to 3 decimal places.)"

We want to calculate the bounds of a 90% confidence interval.

For a 90% CI, the critical value for z is z=1.645.

The sample of men, of size nm=-0.26 has a proportion of pm=0.612.

$p_m=X_m/n_m=318/520=0.612$

The sample 2, of size nw= has a proportion of pw=0.824.

$p_w=X_w/n_w=379/460=0.824$

The difference between proportions is (pm-pw)=-0.212.

$p_d=p_m-p_w=0.612-0.824=-0.212$

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_m+X_w}{n_m+n_w}=\dfrac{318+379}{520+460}=\dfrac{697}{980}=0.711$

The estimated standard error of the difference between means is computed using the formula:

$s_{pm-pw}=\sqrt{\dfrac{p(1-p)}{n_m}+\dfrac{p(1-p)}{n_w}}=\sqrt{\dfrac{0.711*0.289}{520}+\dfrac{0.711*0.289}{460}}\\\\\\s_{pm-pw}=\sqrt{0.00039+0.00045}=\sqrt{0.001}=0.029$

Then, the margin of error is:

$MOE=z \cdot s_{pm-pw}=1.645\cdot 0.029=0.0477$

Then, the lower and upper bounds of the confidence interval are:

$LL=(p_1-p_2)-z\cdot s_{p1-p2} = -0.212-0.0477=-0.260\\\\UL=(p_1-p_2)+z\cdot s_{p1-p2}= -0.212+0.0477=-0.165$

The 90% confidence interval for the difference between proportions is (-0.260, -0.165).

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3 years ago

A local agricultural cooperative claims that 55% of about 60,000 adults in a country believe that gardening should be part of th

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Using the normal distribution and the central limit theorem, it is found that there is a 0.0409 = 4.09% probability that, from a simple random sample of 300 adults in the county, less than 50% would say they believe that gardening should be part of the school curriculum.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.

After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

By the Central Limit Theorem, the sampling distribution of sample proportions for a proportion p in a sample of size n has $\mu = p, s = \sqrt{\frac{p(1 - p)}{n}}$

In this problem:

The proportion is of 55%, hence $p = 0.55$

The sample has 300 adults, hence $n = 300$

Then, the mean and the standard error are given by:

$\mu = p = 0.55$

$s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.55(0.45)}{300}} = 0.0287$

The probability is the p-value of Z when X = 0.5, hence:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.5 - 0.55}{0.0287}$

$Z = -1.74$

$Z = -1.74$ has a p-value of 0.0409.

0.0409 = 4.09% probability that, from a simple random sample of 300 adults in the county, less than 50% would say they believe that gardening should be part of the school curriculum.

A similar problem is given at brainly.com/question/25800303

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